dbb12311-876c-4f57-9480-744abc381c94.html

<p>When you write a quadratic in factored form, that helps you find the \(x\)-intercepts.<br> 
    \(y=x^2+2x-3\)<br>
    \(y=(x+3)(x-1)\)<br>
    </p>
    <p>When we set \(y=0\), we find the places where the parabola crosses the \(x\)-axis.</p>
    
    <p>\(0=(x+3)(x-1)\)</p>
    
    <p>So, \(0=x+3\) and \(0=x-1\)<br>
    \(x=-3\) and 1. </p>
    
    <p>-3 and 1 are called the real solutions to the quadratic equation.</p>
    
    <p>We can also start with the real solutions and use them to write quadratic equations.</p>
    
    <p>When we know that \(x=-3\) and 1, then that means \(0=x+3\) and \(0=x-1\).</p>
    
    <p>So, \(0=(x+3)(x-1)\).</p>
    
    <p>And, using the FOIL method to multiply this out, we come to <br>
    \(0=x^2+2x-3.\)</p>
    
    <p>When \(y\) is not equal to zero, we have the following quadratic equation</p>
    
    <p>\(y=x^2+2x-3\), which is exactly where we started!</p>
  
    <br>
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    <h4>Try It: Writing Quadratic Equations from Real Solutions</h4>
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<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal1" data-schema-version="1.0">
  	<div class="os-raise-ib-cta-content">
		<ol class="os-raise-noindent">
    		 <li>Write a quadratic from the real solutions \(x=1\) and \(x=7\) without graphing.</li>
		</ol>        	
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  	<div class="os-raise-ib-cta-prompt">
  	  <p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
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</div>
<div class="os-raise-ib-content" data-wait-for-event="Reveal1" data-schema-version="1.0">
 <p>Compare your answer: <br>
  \(x^2-8x+7\)</p>
 <p>\(x=1\) and \(x=7\) become \(x-1=0\) and \(x-7=0\). \(y=(x-1)(x-7)= x^2-x-7x+7= x^2-8x+7\).</p>
</div>
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